Question

The number of flaws in bolts of cloth in textile manufacturing is assumed to be Poisson distributed with a mean of 0.1 flaw per square meter. X is a Poisson random variable that denotes the number of flaws in one square meter of cloth with λ = 0.1.
a) What is the probability that there are 2 flaws in 1 square meter?
b) What is the probability that there is one flaw in 10 square meters? Y is a Poisson random variable that denotes the number of flaws in 10 square meters of cloth, with λ = np = 10*0.1 =1.
c) What is the probability that there are no flaws in 20 square meters of cloth?

Answer 1

Since \(X\) has a Poisson distribution, the probability of \(x\) flaws occurring would be
\[p(x)=\frac{e^{-\lambda}\lambda^x}{x!}\]
(a) \(p(2)=\dfrac{e^{-0.1}0.1^2}{2!}\)

(b) \(\lambda\) is given in terms of \(\dfrac{0.1\text{ flaw}}{1\text{ square meter}}\), so you must first change these units to number of flaws per 10 m².
\[\dfrac{0.1\text{ flaw}}{1\text{ square meter}}\times\frac{10}{10}=\dfrac{1\text{ flaw}}{10\text{ square meters}}~~\Rightarrow~~\lambda=1\]
So, \(p(1)=\dfrac{e^{-1}1^1}{1!}\).

(c) Similarly to part (b), you have to adjust \(\lambda\):
\[\dfrac{0.1\text{ flaw}}{1\text{ square meter}}\times\frac{20}{20}=\dfrac{2\text{ flaw}}{20\text{ square meters}}~~\Rightarrow~~\lambda=2\]
So, \(p(0)=\dfrac{e^{-2}2^0}{0!}\).